Let \(X\) and \(Y\) be two independent discrete random variables such that the moment generating functions of \(X\) and \(X+Y\) are given by \[ M_X(t)=\frac{1+2e^{-t}+3e^{2t}}{6},\quad t\in\mathbb{R}, \] and \[ M_{X+Y}(t)=\frac{2+e^{t}+3e^{3t}}{6},\quad t\in\mathbb{R}, \] respectively. Then which of the following statements is correct?
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Read the distribution of X directly off its MGF, then divide the MGF of X+Y by the MGF of X to find the MGF of Y. A constant random variable equal to c has MGF e^{ct}.
Step 1: Split $M_X(t)$ into probability weighted exponentials.
Write $M_X(t)=\frac{1}{6}+\frac{2}{6}e^{-t}+\frac{3}{6}e^{2t}$. Matching each term $p\,e^{xt}$ to a point mass, $X$ is $-1$ with chance $1/3$, $0$ with chance $1/6$, and $2$ with chance $1/2$.
Step 2: Do the same for $M_{X+Y}(t)$.
\[ M_{X+Y}(t)=\frac{2}{6}+\frac{1}{6}e^{t}+\frac{3}{6}e^{3t} \] So $X+Y$ takes the value $0$ with chance $1/3$, $1$ with chance $1/6$, and $3$ with chance $1/2$.
Step 3: Compare the two lists side by side.
$X$ values: $-1,0,2$ with weights $1/3,1/6,1/2$. $X+Y$ values: $0,1,3$ with the same weights $1/3,1/6,1/2$ in the same order. Each value of $X+Y$ is exactly $1$ more than the matching value of $X$, and the weights line up perfectly. This pattern only happens if $Y$ adds a fixed shift of $1$ to every outcome of $X$, that is, $Y$ is the constant $1$.
Step 4: Confirm algebraically.
If $Y\equiv1$, then $M_Y(t)=e^t$, and multiplying, \[ M_X(t)M_Y(t)=e^t\cdot\frac{1+2e^{-t}+3e^{2t}}{6}=\frac{e^t+2+3e^{3t}}{6} \] which is exactly $M_{X+Y}(t)$ as given. So the guess checks out and $Y=1$ with probability $1$, meaning $\mathrm{Var}(Y)=0$.
Step 5: Since $Y$ is the constant $1$, $XY=X$.
So the question reduces to reading probabilities of $X$ directly: \[ P(XY=0)=P(X=0)=\frac{1}{6},\qquad P(XY=2)=P(X=2)=\frac{1}{2} \] This confirms (A) is right and (B) is wrong (it claims $1/6$ but the true value is $1/2$).
Step 6: Check the remaining options quickly.
\[ E(X)=-1\cdot\frac{1}{3}+0\cdot\frac{1}{6}+2\cdot\frac{1}{2}=\frac{2}{3}\neq0 \] so (C) fails, and since $Y$ never varies, $\mathrm{Var}(Y)=0\neq1$, so (D) fails too.
